physics question pertaining to caliber [Archive] - Glock Talk

PDA

View Full Version : physics question pertaining to caliber


OctoberRust
12-28-2011, 17:36
ok so if we had a bullet pushing, say, 500 FT/LBS of energy, that would equate to 500 lbs moved one foot, or at least in physics, right? I understand Newton's third law, however, why isn't the shooter knocked down by 500 ft/lbs at the muzzle if the shooter is 190 lbs? This may seem like a dumb question, but, can anyone explain this more in detail?

cowboy1964
12-28-2011, 18:30
Energy is the not the same as momentum. Momentum is what actually imparts force to an object. Also the force applied by a bullet is over a very short period of time.

The momentum of major handgun service calibers is in the 20-30 ft lb/sec range. And again, that force is applied over a short time (milliseconds basically).

This is the best explanation I can find so far:

http://www.angelfire.com/art/enchanter/terminal.html

Glolt20-91
12-28-2011, 19:21
Because we're talking fluid and thermal dynamics.

chemcmndr
12-28-2011, 20:04
Energy is defined as the "ability to do work." However, you are right with Newton's 3rd Law. The recoil that your body and the firearm absorb during firing will not be greater than the recoil felt by the target when the bullet strikes it. For example, when Myth Busters did the experiment of shooting a pig carcass wearing a Kevlar vest with a 12 gauge slug, the pig only fell off of the hook, instead of being blown backwards like the movie myths sometimes show.

chasbo00
12-28-2011, 21:38
The reason is that the gun's recoil energy does not equal the bullet's muzzle energy. The gun's recoil energy is far less because the gun weighs a whole lot more than the bullet. What are nearly equal are the gun's and bullet's momentum (mass times velocity). I'll use your example of a bullet muzzle energy of 500 ft-lbs and use a 230 grain bullet and a 2-pound gun. The bullet's momentum is the mass of the bullet times its muzzle velocity (about 990 fps in this case based on your 500 ft-lbs of bullet muzzle energy and a 230 grain bullet) and this product equals the mass of the gun times its velocity. Solving for the gun's velocity gives you about 16 fps for the gun's velocity. Plugging this velocity into the energy equation (.5*gun mass*gun's velocity squared) gives you a little more than 8 ft-lbs of gun recoil energy.

Or, if you don't like doing math, here is a recoil calculator:

http://www.huntamerica.com/recoil_calculator/

fredj338
12-28-2011, 22:12
As stated, energy is the ability to do work. A not often thought of aspect of energy, is the bullet is passing thru the target & if it deforms, the bullet is using that energy to deform. You don't see energy being an issue w/ handgun rounds until you get to higher levels IMO, like 44mag, & ONLY if the bullet is designed to take advantage of that energy.
I have shot 4# jack rabbits w/ 250gr LSWC @ 1300fps, a tremendous amount of energy. The rabbit were left unimpressed unless bone was hit. Move to a 357mag/125gr JHP, the bullet can now take advantage of the energy & Mr Bunny stops right there, even though the energy numbers may be half.
I have seen large African game take multiple 6000ft# energy strikes & still come on unimpressed. There have been numerous cases of large male attackers taking 12ga buckshot hits & continue to fight. SO energy is a nice number but unless balanced w/ a projectile that takes advatnage of it, it's almost meaningless.

Tiro Fijo
12-28-2011, 22:42
Here's the scary part: quantum physicists say everything you do or see is actually done in another plane and what you see is merely a reflection of that. :shocked:

OctoberRust
12-29-2011, 10:26
Interesting. So momentum needs to carry that word of KE over to a target. Therefore simply cannot due to the size the projectile lacks from a service pistol?

English
12-29-2011, 12:34
ok so if we had a bullet pushing, say, 500 FT/LBS of energy, that would equate to 500 lbs moved one foot, or at least in physics, right? I understand Newton's third law, however, why isn't the shooter knocked down by 500 ft/lbs at the muzzle if the shooter is 190 lbs? This may seem like a dumb question, but, can anyone explain this more in detail?

Momentum is conserved as momentum and although energy is also conserved it is converted to different forms of energy in any energetic action. i am sure that won't help you very much but read on and it might make more sense.

Momentum is equal to mass times velocity so let us say M = mv
Kinetic energy is equal to half the mass times the square of the velocity, so to avoid using superscript 2s, let us say KE = mvv/2

The first equation tells us thatif we increase the mass by some multiple x and decrease the velocity by dividing it by x the momentum remains the same. That is (xm)(v/x) = mv. To give that sample numbers, we can double the velocity and halve the mass and keep the momentum the same.

Because recoil is a transfer of momentum it is this transfer that tends to push us backwards as we fire. As an example, imagine a 200gn bullet fired at 1,000fps by a 200lb man. The weight of the bullet in lbs is 200/7000, so its momentum is 200x1000/7000. That has to be equal and opposite to the velocity of the man so that momentum is conserved, his velocity has to be the bullet's velocity divided by his mass. The result is that the man moves backwards at 1/7th of a foot per second. When Shane shot the BG who dissapeared backwards through the scenery the bullet might have been 300 grains but its velocity would have been more like 666 fps so the speed he should really have moved at should have been somewhere between 1.5 and 2.5 inches per second.

So, if we double the velocity of the bullet but halve its mass we get the same total recoil but see what happens to the kinetic energy:

KE goes from mvv/2 to (m/2)(2v)(2v)/2 and that is (m/4)(4vv) or mvv. So we have doubled the KE while keeping the same bullet momentum or total recoil.

In short, the KE figure tells you nothing about recoil unless you can use it to work back to momentum from either bullet mass or velocity.

English

G19aps
12-30-2011, 08:31
No hold on just a second. I saw R Lee Ermey on his TV show say that .45 Hardball will knock a man down 9 times out of 10. Is that not true? ;)

English
12-30-2011, 10:12
No hold on just a second. I saw R Lee Ermey on his TV show say that .45 Hardball will knock a man down 9 times out of 10. Is that not true? ;)

People with an indepth knowledge of this subject can only say yes and no.

The latest research, which Ermey might not have seen and which is only a continuation of the previous research as more volunteers are found, gives a figure of 8.7 out of 10 to one decimal place. As there are relatively few volunteers this number can be considered only approximate, so Ermeys use of 9 might be considered a reasonable approximation. this would be a tentative yes.

Further, the 8.7 number is complicated by the self selection involved in using volunteers. The momentum transfer of the bullet moves the person shot backwards at around 2 inches per second and in order not to be knocked down the person has to take a step back to regain his ballance. There lies the problem. We can assume that some people are incapacitated and unable to step back. Other can't work out that they need to step back and so are knocked over- rather slowly it must be said. Unfortunately the nature of the volunteers for this experiment means that many fall into the latter category and so it is hard to say what proportion of the 8.7 in a normal population would fall over. Best estimates are approximately 1.5. Such are the problems of experimental science and the scientists are working to refine their experimental model. In this fuller sense, Ermey would be incorrect and should have said one or two but even this figue is complicated by those who don't step back but could if they worked out what foot to use.

When this experiment was repeated with rats, none of which volunteered, 9.5 out of ten were knocked over. When repeated with deer, which also had not volunteered, 10 out of 10 ran away.

English

fredj338
12-30-2011, 11:59
No hold on just a second. I saw R Lee Ermey on his TV show say that .45 Hardball will knock a man down 9 times out of 10. Is that not true? ;)

I know you are tounge in cheek w/ that but the issue w/ momentum is it looks good on small/light targets. You see it shooting steel or bowling pins, the heavier/slower bullet offers more time on the target & greater force so the target falls easier. Apply the same thing to a 200# steel target or person, nothing happens w/ either 9mm or 45. Even higher energy rifle rounds wont knock down a 200# target w/ certainty.:dunno: When you get to really high energy numbers, like the Afircan big bores, then you start to see "light" targets, like 200# steel, go down pretty quickly. Then again, the recoil from some of those large African/6000 ft# rounds, can actually move a 200# shooter quite a bit.

English
12-30-2011, 12:37
I know you are tounge in cheek w/ that but the issue w/ momentum is it looks good on small/light targets. You see it shooting steel or bowling pins, the heavier/slower bullet offers more time on the target & greater force so the target falls easier. Apply the same thing to a 200# steel target or person, nothing happens w/ either 9mm or 45. Even higher energy rifle rounds wont knock down a 200# target w/ certainty.:dunno: When you get to really high energy numbers, like the Afircan big bores, then you start to see "light" targets, like 200# steel, go down pretty quickly. Then again, the recoil from some of those large African/6000 ft# rounds, can actually move a 200# shooter quite a bit.

Fred,
I admire your knowledge and experience, but the effect of bullets on steel plates or bowling pins has little relationship to their effect on humans.

What knocks a steel plate or bowling pin over is transfer of momentum producing speed of movement of the target. Maximum speed of movement will occur when the bullet bounces back from the plate (I won't bother to keep writin "or pin"). The net momentum must remain the same after the impact and so the negative value of the bouncing bullet must be added to the momentum of the plate, and as that is lighter than it would be if the bullet "stuck" to it, the plate will move back considerably faster.

The difference between a higher energy bullet and a lower energy bullet of the same momentum is what happens to the extra energy. The answer is that the high energy bullet is more likely to splatter sideways - the momentum of the bullet/plate system remains the same because the sideways splatter is approximately ballanced at right angles to the bullet path. This reduces the bullet bounce back momentum and so the plate moves with less velocity than with the slower bullet.

But as we have seen, the man shot is a lot heavier than a plate and only moves back at a combined 2 inches per second or so. If he was in a state to keep shooting normally he could fire some four or five shots by the time he has moved 2 inches and so that would not affect him very much. More important than that is that his recoil is initially localised. Whatever part of him is hit is pushed back faster than 2 inches per second and that distorts his body as the rest of him catches up. That flaps his arms around a bit and knocks him off aim. I actually think this is a useful part of what is achieved when you shoot someone who is trying to shoot you - it is likely to make him miss with the shot he was in the process of firing and take up more of his time to re aim the next but the simple transfer of momentum does him very little real damage. The real damage is done by the transfer of energy.

The energy transfer does two or possibly three short term things. First it puts a big chunk of tissue out of commission and some of that makes it harder to use his body and certainly that takes him extra time to compensate for the lack of muscle or bone. Second, the temporary cavity also jerks him around and is another disruption to his aiming process. Third, if you have the right ammunition and if you are lucky the balistic pressure wave effect will disorient him or make him loose conscious control for anything from a brief to an extended period. In comparisson, momentum transfer is a minor hindrance to him shooting you.

English

Japle
12-30-2011, 12:45
I understand Newton's third law, however, why isn't the shooter knocked down by 500 ft/lbs at the muzzle if the shooter is 190 lbs? This may seem like a dumb question, but, can anyone explain this more in detail?
There's not enough time.

Here's an example: You can move a 2,000 lb safe door with one finger. The door will move slowly, but it'll move. If you punch the door as hard as you can, it won't move at all. In order to move something, the force has to be applied over enough time to overcome the object's inertia.

A gun only pushes on your hand for a few thousandths of a second at most. That's enough to move the gun and your hand and arm, but not enough to move your body.